Determine the value of
\[2002 + \frac{1}{2} \left( 2001 + \frac{1}{2} \left( 2000 + \dots + \frac{1}{2} \left( 3 + \frac{1}{2} \cdot 2 \right) \right) \dotsb \right).\]
Explanation: Let
\begin{align*}
S &= 2002 + \frac{1}{2} \left( 2001 + \frac{1}{2} \left( 2000 + \dots + \frac{1}{2} \left( 3 + \frac{1}{2} \cdot 2 \right) \right) \dotsb \right) \\
&= 2002 + \frac{2001}{2} + \frac{2000}{2^2} + \dots + \frac{3}{2^{1999}} + \frac{2}{2^{2000}}.
\end{align*}Then
\[2S = 2 \cdot 2002 + 2001 + \frac{2000}{2} + \dots + \frac{3}{2^{1998}} + \frac{2}{2^{1999}}.\]Subtracting these equations, we get
\begin{align*}
S &= 4004 - 1 - \frac{1}{2} - \frac{1}{2^2} - \dots - \frac{1}{2^{1999}} - \frac{2}{2^{2000}} \\
&= 4004 - 1 - \frac{1}{2} - \frac{1}{2^2} - \dots - \frac{1}{2^{1999}} - \frac{1}{2^{1999}} \\
&= 4004 - \frac{1}{2^{1999}} (2^{1999} + 2^{1998} + \dots + 2 + 1 + 1) \\
&= 4004 - \frac{1}{2^{1999}} \cdot 2^{2000} \\
&= 4004 - 2 = \boxed{4002}.
\end{align*}